Chemistry: Post your doubts here!

Messages
30
Reaction score
11
Points
18
Can someone explain why the answer is C ?
If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps:)
 
Messages
109
Reaction score
215
Points
53
If you draw the propene molecule, it will help. so the molecule looks something like this: CH2=CH-CH3. Now we have Br2 which is (i) Br-Br and HBr which is (ii) H-Br. In case (i) the nucleophile will attack. the double bond will break and one extra bond will be available on C1 and one more on C2. a Br ion can attach to each one of these free bonds, thus forming option D. in (ii) H can bond to first free bond or the second. Br will attach to the one left. Through this, the process forms option A and B.

Option C is impossible because it implies that both the bromine ions went to the first carbon, but that is not possible because if you revisit molecular bonding, a pi bond is basically an unhybridised p orbital from each atom interacting. SO the breaking of a double bond will automatically free up one unhybridised p orbital on either C atom. You just can't have option C. Others are possible, as described above.

Hope this helps:)
Thank you so much!!!!
 
Messages
30
Reaction score
11
Points
18
Can anyone pls explain why is the is answer C? Thank you :D
So the graph is characteristics for an ENDO thermic reaction which is deltaH +ve value. For the reverse reaction, energy needs to be supplied to increase the potential energy of products to the energy level of intermediate(which has a greater potential energy). So this also requires an INPUT of energy (i.e, +ve sign). This makes C the right answer.
 
Messages
436
Reaction score
236
Points
53
Is there any whatsapp group for this group and if yes can someone kindly add me over there
Should I tell my whatsapp number
 
Messages
2
Reaction score
1
Points
1
Hi can any one explain this??
You have 100.000g of a 60% by mass sulfuric acid mixture of a calculated density of 1.370gcm–3. Calculate the mass of water you would need to add to the 100.000g in order to give a final calculated density of 1.154 g cm–3.?
 
Messages
2
Reaction score
0
Points
1
OK so first you need to know the reactions of the period 3 oxides with water. AL2O3 does not react so optionA b are wrong. Next we react 2NaOH (which is a basic oxide and the product of reaction between oxide and H2O) with corresponding acidic oxide & balance the equations.
NOTE The Q states"neutralised by exactly 1mol....so D is correct as it balances perfectly with 2NaOH.
Al2O3 does react with acid and alkali since it is amphoteric.
a) when Y is added to water H3PO4 will form and if you balance the equation when reacted with Al2O3, you wont get one mole for each of the reactants.
b) when SO3 is added to water H2SO4 will form and if you balance the equation when reacted with Al2O3, you wont get one mole for each of the reactants.
c) same goes to when Na2O reacts with H2SO4. (wont get one mole)
d) the equation is Na2O + H2SO4 = Na2SO4 + H2O. X and Y both have one mole.
Therefore the ans is D.
 
Messages
1
Reaction score
0
Points
1
compressing at constant temp will never liquify any gas... you have to cool the cylinder as well for liquifying... else the cylinder will blast away....
Hey brother, I don’t think this statement is correct. Gas can be liquified by increasing pressure at constant T. Just look at phase diagram.
Reason 2 is wrong is bcause using the relationship of PV=PV, volume of gas at 6000k pa should be 19cm3.
However, the measure volume is 20.5cm3.
This shows that the gas is less compressible than expected (what we call the positive deviation from ideal gas behaviour)
Option 2 cannot be valid. Gas can be liquify if they are compressible i.e. stronger intermolecular attraction force (a negative deviation from ideal gas behavior should be observed.
Meaning volume should be LOWER than ideality when gas is liquified not higher.
 
Messages
3
Reaction score
0
Points
1
Which solid contains more than one kind of bonding?
A. Iodine
B. Silicon dioxide
C. Sodium chloride
D. Zinc

The answer is A but I don't understand really why. I know that it has covalent bonds and van der waals forces, but don't silicon dioxide and sodium chloride also have those?

Which element is likely to have an electronegativity similar to that of aluminium?
A. barium
B. beryllium
C. magnesium
D. strontium
I have no clue as to how to figure this one out

Suggest an explanation for the existence of IF7 and for the non-existence of ClF7? The answer says because I is a bigger atom than Cl but I need a more thorough explanation as I didn't understand anything from that answer.

Thank you :)
silicon dioxide has a giant molecular structure and consist of only strong covalent bomd
 
Messages
2
Reaction score
0
Points
1
in an experiment a sample of a pure gas is put into a gas syringe at a temperature of 300K and pressure of 16 kpa. the gas is compressed until the volume occupied by the gas is halved.
after the compression, the temperature of the gas in syringe is 375 and the pressure is 40 kpa.
is the gas is ideal
The gas is behaving ideally, use PV=nRT before and after compression and make n the subject. You'll see that for both equations n is constant.
 
Top